To determine the enthalpy change for the reaction:
[tex]\[ \text{C}_2\text{H}_4 + \text{H}_2\text{O} \rightarrow \text{C}_2\text{H}_5\text{OH} \][/tex]
we use the enthalpies of formation ([tex]\( \Delta H_f^\circ \)[/tex]) of the reactants and products provided:
- Enthalpy of formation of Ethylene ([tex]\( \text{C}_2\text{H}_4 \)[/tex]): 52.26 kJ/mol
- Enthalpy of formation of Water ([tex]\( \text{H}_2\text{O} \)[/tex]): -285.83 kJ/mol
- Enthalpy of formation of Ethanol ([tex]\( \text{C}_2\text{H}_5\text{OH} \)[/tex]): -277.69 kJ/mol
The enthalpy change ([tex]\( \Delta H \)[/tex]) for the reaction is calculated by subtracting the sum of the enthalpies of formation of the reactants from the sum of the enthalpies of formation of the products:
[tex]\[ \Delta H = \sum \Delta H_f^\circ (\text{products}) - \sum \Delta H_f^\circ (\text{reactants}) \][/tex]
Now, let us break this down step-by-step:
1. Sum of the enthalpies of formation of the products:
In this reaction, the product is Ethanol ([tex]\( \text{C}_2\text{H}_5\text{OH} \)[/tex]).
[tex]\[ \sum \Delta H_f^\circ (\text{products}) = \Delta H_f^\circ (\text{C}_2\text{H}_5\text{OH}) = -277.69 \, \text{kJ/mol} \][/tex]
2. Sum of the enthalpies of formation of the reactants:
The reactants are Ethylene ([tex]\( \text{C}_2\text{H}_4 \)[/tex]) and Water ([tex]\( \text{H}_2\text{O} \)[/tex]).
[tex]\[ \sum \Delta H_f^\circ (\text{reactants}) = \Delta H_f^\circ (\text{C}_2\text{H}_4) + \Delta H_f^\circ (\text{H}_2\text{O}) \][/tex]
[tex]\[ \sum \Delta H_f^\circ (\text{reactants}) = 52.26 \, \text{kJ/mol} + (-285.83 \, \text{kJ/mol}) \][/tex]
[tex]\[ \sum \Delta H_f^\circ (\text{reactants}) = 52.26 \, \text{kJ/mol} - 285.83 \, \text{kJ/mol} \][/tex]
[tex]\[ \sum \Delta H_f^\circ (\text{reactants}) = -233.57 \, \text{kJ/mol} \][/tex]
3. Calculate the enthalpy change of the reaction:
[tex]\[ \Delta H = \sum \Delta H_f^\circ (\text{products}) - \sum \Delta H_f^\circ (\text{reactants}) \][/tex]
[tex]\[ \Delta H = -277.69 \, \text{kJ/mol} - (-233.57 \, \text{kJ/mol}) \][/tex]
[tex]\[ \Delta H = -277.69 \, \text{kJ/mol} + 233.57 \, \text{kJ/mol} \][/tex]
[tex]\[ \Delta H = -44.12 \, \text{kJ/mol} \][/tex]
Thus, the enthalpy change for the reaction [tex]\(\text{C}_2\text{H}_4 + \text{H}_2\text{O} \rightarrow \text{C}_2\text{H}_5\text{OH}\)[/tex] is [tex]\(-44.12 \, \text{kJ/mol}\)[/tex].
